Show that the sum of an AP whose first term is $ a $, the second term $ b $ and the last term $ c $, is equal to $ \frac{(a+c)(b+c-2 a)}{2(b-a)} $
Given:
An AP whose first term is \( a \), the second term \( b \) and the last term is \( c \).
To do:
We have to show that the sum of an AP whose first term is \( a \), the second term \( b \) and the last term \( c \), is equal to \( \frac{(a+c)(b+c-2 a)}{2(b-a)} \).
Solution:
In the given AP,
First term $=a$
Common difference $=b-a$
Last term $a_{n}=c$
$a_{n}=l=a+(n-1) d$
$c=a+(n-1)(b-a)$
$c-a=(n-1)(b-a)$
$(n-1)=\frac{c-a}{b-a}$
$n=\frac{c-a}{b-a}+1$
$n=\frac{c-a+b-a}{b-a}$
$n=\frac{c+b-2 a}{b-a}$
Sum of the $n$ terms of an AP $S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$=\frac{(b+c-2 a)}{2(b-a)}\left[2 a+\left\{\frac{b+c-2 a}{b-a}-1\right\}(b-a)\right]$
$=\frac{(b+c-2 a)}{2(b-a)}\left[2 a+\frac{c-a}{b-a} (b-a)\right]$
$=\frac{(b+c-2 a)}{2(b-a)}(2 a+c-a)$
$=\frac{(b+c-2 a)}{2(b-a)} (a+c)$
Hence proved.
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